# Calculate the sum of inverse values of ${n\choose 0}, {n\choose 1}, … {n\choose n}$

Calculate $$A={1\over {n\choose 0}}+ {1\over {n\choose 1}}+ ...+{1\over {n\choose n}}$$

and

$$B={1\over {n\choose 0}}- {1\over {n\choose 1}}+ ...+{(-1)^n\over {n\choose n}}$$

My idea for $A$ is some probabilistic reasoning. Color the sets $$\{\}, \{1\}, \{1,2\}, \{1,2,3\},...\{1,2,...,n\}$$ and ask our self what is the probability that I choose colored set among all sets. Clearly this is exactly ${n+1\over 2^n}$ and on the other side it is $A$:

probability that I take empty set is ${1\over {n\choose 0}}$,

probability that I take colored set with 1 element is ${1\over {n\choose 1}}$

probability that I take colored set with 2 elements is ${1\over {n\choose 2}}$

and so on ...

So $A ={n+1\over 2^n}$. But I have no idea how to attack $B$.

• ${n\choose k}={n\choose n-k}$, so when $n$ is odd and we have $(-1)^k\neq (-1)^{n-k}$, lots of terms cancel. Could that help? – Mastrem Oct 18 '17 at 19:05
• The first sum is certainly larger than $2$ for $n>1$. So I don't see how that sum could directly represent some probability. And furthermore, $A$ can't equal $\frac{n+1}{2^n}$ as you conclude, because it is typically smaller than $2$. – alex.jordan Oct 18 '17 at 19:08
• So I blow it, do you have an idea how to fix it? – Aqua Oct 18 '17 at 19:16
• A Google search for "sum of reciprocals of binomial coefficients" turns up many relevant results. There seems to be no closed form, but a number of equivalent sums and a recurrence are there. – marty cohen Oct 18 '17 at 19:17

A good moment for exploiting the beauty of Euler's Beta function. $$\begin{eqnarray*}\sum_{k=0}^{n}\frac{(-1)^k}{\binom{n}{k}}&=&(n+1)\sum_{k=0}^{n}(-1)^k \frac{\Gamma(k+1)\Gamma(n-k+1)}{\Gamma(n+2)}\\&=&(n+1)\sum_{k=0}^{n}(-1)^kB(k+1,n-k+1)\\&=&(n+1)\int_{0}^{1}\sum_{k=0}^{n}(-1)^k (1-x)^k x^{n-k}\,dx\\&=&(n+1)\int_{0}^{1}(-1)^n(1-x)^{n+1}+x^{n+1}\,dx\\&=&\color{blue}{\frac{n+1}{n+2}((-1)^n+1)}.\end{eqnarray*}$$
I found the solution, though it is not mine. Write $${n\choose k}' := {(-1)^k\over {n\choose k}}$$
Then it is easy to prove $${n\choose k+1}'-{n\choose k}' =-{n+1\over n}{n-1\choose k}'$$ so we have: \begin{eqnarray} B&=&\sum _{k=0}^n {n\choose k}' \\&=&-{n+1\over n+2} \sum _{k=0}^n \Big[ {n+1\choose k+1}'-{n+1\choose k}'\Big] \\ &=& -{n+1\over n+2}\Big[ {n+1\choose n+1}'-{n+1\choose 0}'\Big] \\ &=& -{n+1\over n+2}\big( (-1)^{n+1}-1\big)\\ &=& {n+1\over n+2}\big( (-1)^n+1\big)\\ \end{eqnarray}